Why is $x^{1/n}$ continuous?

Question

Why is $x^{1/n}$ continuous for positive $x,n$ where $n$ is an integer? I can't see how it follows from the definition of limit. And I don't see any suitable inequalities so is this an application of Bernoulli's or Jensen's inequality?

Answer 1

The idea is that $f(x) = x^{1/n}$ is Lipschitz in a small neighborhood of any point $x > 0$.

We need the following estimate. For $x, y > 0$, denote $u := f(x)$ and $v := f(y)$, so that $x = u^n$ and $y = v^n$. Then we can bound $|f(x)-f(y)|$ by $$ \begin{align*} |f(x) - f(y)| = |u-v| &= \left| \frac{x-y}{u^{n-1} + u^{n-2} v + \cdots + u v^{n-2} + v^{n-1}} \right| \\ &\leqslant \frac{|x-y|}{u^{n-1}} = \frac{|x-y|}{x^{\frac{n-1}{n}}}. \end{align*} $$ Therefore, for any $a > 0$, $f$ is $\ell$-Lipschitz in the neighborhood $\left[\frac{a}{2}, 2a \right]$ of $a$ for $\ell = \ell(a) = \left( \frac{a}{2} \right)^{- \frac{n-1}{n}}$. Continuity at $a$ follows from this. $\qquad \diamond$

---

Remarks.

1. In fact, $f$ is also differentiable at every $a > 0$. Moreover, it is possible to strengthen the above proof to obtain the derivative as well.

2. It turns out the function is Hölder continuous on every interval $[0, a]$, though it is not Lipschitz (assuming $n > 1$). Therefore, $f$ is continuous but not differentiable at the origin.

3. While Mariano's answer explicitly uses that $f(x)$ is the inverse function to $g(x) = x^n$, we also exploit this fact -- but implicitly. Indeed, since $g$ is invertible and has nonzero derivative at every point in $[0, \infty)$, the inverse function theorem guarantees that $f$ is also differentiable at every $a > 0$.

Answer 2

One way to show this is to observe that it is the inverse function of $x\mapsto x^n$, which is itself continuous.